Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $n \neq 0$. $t = \dfrac{8(2n - 5)}{-10} \times \dfrac{4n}{18n - 45} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $t = \dfrac{ 8(2n - 5) \times 4n } { -10 \times (18n - 45) } $ $ t = \dfrac {4n \times 8(2n - 5)} {-10 \times 9(2n - 5)} $ $ t = \dfrac{32n(2n - 5)}{-90(2n - 5)} $ We can cancel the $2n - 5$ so long as $2n - 5 \neq 0$ Therefore $n \neq \dfrac{5}{2}$ $t = \dfrac{32n \cancel{(2n - 5})}{-90 \cancel{(2n - 5)}} = -\dfrac{32n}{90} = -\dfrac{16n}{45} $